A) 6.6 N, 35 \[m{{s}^{-1}}\]
B) 6 N, 37 \[m{{s}^{-1}}\]
C) 7.5 N, 46 \[m{{s}^{-1}}\]
D) 8 N, 38 \[m{{s}^{-1}}\]
Correct Answer: A
Solution :
Frequency of revolution of stone, \[f=40\,rev/\min =\frac{40}{60}rev/s\] Angular speed of the stone, \[\omega =2\pi f\] \[=2\pi \times \frac{40}{60}=\frac{4\pi }{3}\]rad \[{{s}^{-1}}\] The centripetal force is provided by the tension (T) in the string i.e. \[T=\frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}\] \[=0.25\times 1.5\times {{\left( \frac{4\pi }{3} \right)}^{2}}N=6.48\,N\approx 6.6\,N\] As the string can withstand a maximum tension of 200 N. \[\therefore \] \[{{T}_{\max }}=\frac{mv_{\max }^{2}}{r}\] \[\Rightarrow \] \[{{V}_{\max }}=\sqrt{\frac{r{{T}_{\max }}}{m}}=\sqrt{\frac{1.5\times 200}{0.25}}\] \[=34.64\,m{{s}^{-1}}\approx 35\,m{{s}^{-1}}\] \[T=6.6N,\,{{v}_{\max }}=35\,m{{s}^{-1}}\]
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